Limit of piecewise function / lim x-> 3 of 1/(x-3)^2

[paulxzt]

Can someone check/help me with some review problems I did:

1) lim x-> 3 of 1/(x-3)^2 = is this DNE ?

2) lim x->4 for
f(x) =
{ x-3 , x not equal to 4
{ 5 , x=4

do I begin by plugging in 4 for the equations? a jump at 4?

3) lim x ->infinity for
4sin(3x) / x
will this be 0 ?

 

[arthur ohlsten]

1)
lim x-->3 of 1/[x-3]^2 = +infinity regardless of how you approach 3, from the right or left

2)
lim x-->4 of f[x] where:
f[x]=x-3 x not =4
f[x]=5 x=4
f[x] is a straight line with slope of 1 and y intercept of -3
f[x] has a "hole" in the line and a value of 5 at x=4
a jump at 4, and is discontinuous at x=4

3) lim x-->infinity of f[x]
f[x]=4 sin(3x) /x

4 sin3x is bounded by +4 and -4
4sin[3x] /x=0 as x-->infinity

hope I am right.
If I am wrong, some one should correct the answer
Arthur

 

[shortman313]

We have this dang limit problem due tomorrow and I am stumped to a t with it. Could someone please give me some guidance. Thanks a lot.

lim
x-->0 (Sqrt(1+tan x) - Sqrt(1+sin x)) all over x^3

 

[Count Iblis]

We have this dang limit problem due tomorrow and I am stumped to a t with it. Could someone please give me some guidance. Thanks a lot.

lim
x-->0 (Sqrt(1+tan x) - Sqrt(1+sin x)) all over x^3

Use that Sqrt[1+x] = 1+ 1/2 x + 1/2 (-1/2)/2 x^2 + 1/2(-1/2)(-3/2)/6 x^3 +
O(x^4)

You can thus write the numerator as:

1/2 [tan(x) - sin(x)] - 1/8 [tan^2(x) - sin^2(x)] + 1/16 [tan^3(x) - sin^3(x)] +
O(x^4)

Note that sin(x) and tan(x) are of order x near x = 0 so that the neglected terms are of order x^4 and won't contribute to the limit. The next step is to insert the series expansions of the sin and the tan. The series expansion of the tan can be obtained by division. We only need to work to order x^3, so:

tan(x) = sin(x)/cos(x) = (x - x^3/6)/(1-x^2/2) + O(x^5) =

(x - x^3/6)(1 + x^2/2) + O(x^5) =

x + x^3/3 + O(x^5)

So, you see that in the first square brackets you get an term of order x^3 and you can divide by x^3 to get a finite limit for x --> 0 (I find 1/12). The third square brackets yields a term of order x^5 and thus also doesn't contribute to the limit.

The term in the second square brackets is of order x^4 and thus yields zero when divided by x^3 and x -->0.